Question

A particle starts from rest and has an acceleration of $2$ $m{s^{ - 2}}$ for $10$ $s$. After that, the particle travels for $30$ $s$ with constant speed and then undergoes a retardation of $4$ $m{s^{ - 2}}$ and comes back to rest. The total distance covered by the particle is
(A) $650$ $m$
(B) $700$ $m$
(C) $750$ $m$
(D) $800$ $m$

Answer 1

In order to solve this question, split the question in three parts. The first part will consist of the motion of the particle for the first $10$ $s$. The second part will consist of the motion of the particle for the next $30$ $s$. And the third part will consist of the rest of the motion.
Formulae to be used are: $s = ut + \dfrac{1}{2}a{t^2}$
${v^2} = {u^2} + 2as$
$v = u + at$
Where $s$ is the distance travelled, $a$ is the acceleration, $u$ and $v$ are the initial and final velocities respectively and $t$ is the time taken.

Complete step by step answer:
1. The distance travelled in first part:
Since the particle starts from rest, the initial velocity will be zero.
$u = 0$
The acceleration of the particle is $a = 2$ $m{s^{ - 2}}$.
The time taken in this part is $t = 10$ $s$.
Therefore, the distance covered in first part will be
${s_1} = ut + \dfrac{1}{2}a{t^2} \\\ {s_1} = (0)(10) + \dfrac{1}{2}(2){(10)^2} \\\$
$\Rightarrow {s_1} = 100$ $m$.

2. The distance travelled in second part:
Since the particle was previously in motion, here the initial velocity will not be zero, instead it will be $v = u + at$. Here the value of $u$, $a$ and $t$ are from the first part.
$\Rightarrow v = 0 + (2)(10)$
$\Rightarrow v = 20$ $m{s^{ - 1}}$.
Now this $v = 20$ $m{s^{ - 1}}$ will be the initial velocity in the second part.
The acceleration of the particle is $a = 0$ $m{s^{ - 2}}$.
The time taken in this part is $t = 30$ $s$.
Therefore, the distance covered in second part will be
${s_2} = ut + \dfrac{1}{2}a{t^2} \\\ {s_2} = (20)(30) + \dfrac{1}{2}(0){(30)^2} \\\$
${s_2} = 600$ $m$.

3. The distance travelled in third part:
Since the particle was previously in motion with a constant velocity, here the initial velocity will be $20$ $m{s^{ - 1}}$. The final velocity will be zero, as it comes to rest.
The retardation of the particle is $4$ $m{s^{ - 2}}$, therefore the acceleration will be $- 4$ $m{s^{ - 2}}$.
The time taken by the particle to come to rest will be
$v = u + at \\\ \Rightarrow t = \dfrac{{v - u}}{a} \\\$
$\Rightarrow t = \dfrac{{0 - 20}}{{ - 4}}$
$\Rightarrow t = 5$ $s$.

Therefore, the distance covered in third part will be
${s_3} = ut + \dfrac{1}{2}a{t^2} \\\ {s_3} = (20)(5) + \dfrac{1}{2}( - 4){(5)^2} \\\$
${s_3} = 50$ $m$.
Therefore, the net distance travelled by the particle will be ${s_1} + {s_2} + {s_3}$ which will be $750$ $m$.
Hence, the total distance covered by the particle is $750$ $m$.

So, the correct answer is “Option C”.

Note:
Remember the kinematics equations of motion, this will help you to solve various problems. In problems where the displacement, velocity and acceleration are given in different intervals of time, split the whole motion in intervals and try to find the values of initial and final position, velocity and acceleration. In this type of problems which is discussed above, always remember to keep the track of the initial and final velocities while jumping from one part to the other, as the particle will be carrying the final velocity from the previous part to the next part as the initial velocity.